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GATE CSE 2015 | Set 3 | Question: 36

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Two hosts are connected via a packet switch with $10^7$ bits per second links. Each link has a propagation delay of $20$ microseconds. The switch begins forwarding a packet $35$ microseconds after it receives the same. If $10000$ bits of data are to be transmitted between the two hosts using a packet size of $5000$ bits, the time elapsed between the transmission of the first bit of data and the reception of the last bit of the data in microseconds is ______.

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2 2 votes
First packet takes all the way:-

$500 \mu s+20\mu s+35 \mu s+500\mu s + 20 \mu s=1075\mu s$

 

Second packet rolls in pipelined fashion. So just consider $T_t$

=> $500 \mu s$

Total: $1575 \mu s$
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1 1 vote
There are 10000/5000 = 2 packets.

Transmission time for 1 packet = 5000 bits / 10^7 bps = 500 micro sec.

Packet 1 reaches in 500 (transmission from source) + 20 ( propagation to switch) +35 ( switch delay) + 500 (transmission from switch) + 20 (propagation to receiver) = 1075 micro sec.

Since packet 2 starts 500 micro sec after packet 1, the last bit of it reaches in 1075 + 500 = 1575 micro sec.

Correct me if I am wrong.
1 1 vote

It represents the complete scenario given in the question.

The basic idea behind this question is that the two packets are travelling parallely .

1 1 vote
Solution GATE 2015 Set 3 paper
GATE CSE 2015 Set 3 | Question: 36

I thought of the time diagram as a tabular representation. And this approach seemed intuitive and easier for me. 

Here, we can add other delays as well if present like queuing and processing ones at each node.

This looks practical because we had just two packets in this question but if we have like many more packets then this approach could be troublesome. There only the Go Classes way of analysis will help and yeah we can make use of these types of tables there as well. 

Hope it helps someone :-)

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